# What are Fourier transforms of optics?

1. Mar 12, 2014

### Choisai

So I'm currently busy studying a Digital Micromirror Device which is used for top-hat beam generation. Programming the input pattern and error diffusion needed for optimal top-hat generation is heavily based on Fourier Optics.

The problem however is: I don't know Fourier optics. I know this is a very broad subject on which entire books are written and that no answer in this forum could ever touch on all its applications or have the neccesary depth, but since this is the internet I would like to hear some explanation of it.

I want to use the error diffusion as described in this paper: http://www.opticsinfobase.org/josab/abstract.cfm?uri=josab-24-6-1268

It describes a telescope construction with a pinhole for spatial filtering. It describes the following:

It is assumed that coherent beam with constant intensity I0 = |E0|² is incident on a transmission mask, which is relay imaged to the image plane, following Fig 2. A fourier plane in the imaging system can be used for Fourier filtering. An example of such system is a two lens system, with lenses of identical focal length, for which Fourier filtering can be performed with a pinhole at the Fourier plane of the first lens. For the sake of simplicity, we assume an imaging system with magnification equal to 1. The electric field after the binary mask with transmission s(x,y) is E(x,y)= E0 X$\bar{s}$(u,v), where $\bar{s}$ is the Fourier transform of S. This field is filtered by a transmission filter p, leading to the field:"
E0 X $\bar{s}$(u,v) X p(u,v).

The resulting field at an image plane can be written as a convolution E'=E0 X $\bar{s}$(u,v)$\otimes$$\bar{p}$. As the convolution with the Fourier transform of the filter $\bar{p}$ acts like a local averaging operation on the elctric field of light after the shpaer, the intensity of the output field at a given point (x,y) is proportional to the square of the average value of s around this point. This is important when a beam shaper before filtering must be designed to be equal to the square root of the target intensity transmission after filtering. The averaging operation provided by the filter in the far field is the key point in obtaining a smooth continuous intensity from a binary pixelated mask.

So my questions are:
1) what are Fourier optics?
2) What is the Fourier plane
3) what are they describing in that paper?
4) What is Fourier filtering
5) What more should I know about Fourier optics and perhaps also what they are describing there?

#### Attached Files:

• ###### Error diffusion paper Figure 2.png
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2. Mar 12, 2014

### UltrafastPED

3. Mar 12, 2014

### my2cts

4. Mar 12, 2014

### Choisai

It's an internship

5. Mar 13, 2014

### Cthugha

From what you describe, it sounds like you might want to read about what the Gerchberg-Saxton algorithm does: http://en.wikipedia.org/wiki/Gerchberg–Saxton_algorithm.

6. Mar 13, 2014

### Choisai

Thank you! It explained a lot of what I wanted to know. But I was wondering: the Fourier transform of a lens looks an awful lot like diffraction of a wave that encounters a split. Is this just coincidence or is Fourier transform and diffraction two sides of the same coin? Diffraction has multiple 'dots' where as the Fourier transform of a single frequency only has two dots and one central dot, but I can't help but wonder if they are more or less the same thing.

Also, on the website they describe band-pass filtering. Is a pinhole an example of such a filter?

7. Mar 13, 2014

### my2cts

I think you should study the basic principles of diffraction. Look for a crash course on the web.
Th knowledge you require cannot be obtained by a question and answering process.

8. Mar 13, 2014

### sophiecentaur

When light passes through any aperture, you will get a diffraction pattern, with or without a piece of round glass (lens) in the hole. The effect of a lens can be shown to be a Fourier transform. Ergo, the two approaches must be equivalent.
A closely related topic is the Zone Plate, which can form an image, as a lens does. See this link.